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  • A1011World Cup Betting

    With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excited as the best players from the best teams doing battles for the World Cup trophy in South Africa. Similarly, football betting fans were putting their money where their mouths were, by laying all manner of World Cup bets.

    Chinese Football Lottery provided a "Triple Winning" game. The rule of winning was simple: first select any three of the games. Then for each selected game, bet on one of the three possible results -- namely W for win, T for tie, and Lfor lose. There was an odd assigned to each result. The winner's odd would be the product of the three odds times 65%.

    For example, 3 games' odds are given as the following:

     W    T    L
    1.1  2.5  1.7
    1.2  3.1  1.6
    4.1  1.2  1.1
    
     

    To obtain the maximum profit, one must buy W for the 3rd game, T for the 2nd game, and T for the 1st game. If each bet takes 2 yuans, then the maximum profit would be ( yuans (accurate up to 2 decimal places).

    Input Specification:

    Each input file contains one test case. Each case contains the betting information of 3 games. Each game occupies a line with three distinct odds corresponding to WT and L.

    Output Specification:

    For each test case, print in one line the best bet of each game, and the maximum profit accurate up to 2 decimal places. The characters and the number must be separated by one space.

    Sample Input:

    1.1 2.5 1.7
    1.2 3.1 1.6
    4.1 1.2 1.1
    
     

    Sample Output:

    T T W 39.31

    思路:

    •开a[3]在每一组的输入中找到最大的一个数,并记录此时的下标,并做pro*=a[maxi]运算;

    •利用mp进行映射将下标输出成字符。

     1 #include <iostream>
     2 using namespace std;
     3 int main() {
     4     double a[3] = { 0 },pro=1.0;
     5     int b[3] = { 0 },maxi;
     6     for (int i = 0; i < 3; i++) {
     7         cin >> a[0] >> a[1] >> a[2];
     8         maxi = a[0] >= a[1] ? 0 : 1;
     9         maxi = a[maxi] >= a[2] ? maxi : 2;
    10         pro *= a[maxi];
    11         b[i] = maxi;
    12     }
    13     char mp[3] = { 'W','T','L' };
    14     for (int i = 0; i < 3; i++) {
    15         printf("%c ", mp[b[i]]);
    16     }
    17     pro = (pro*0.65 - 1) * 2;
    18     printf("%.2lf", pro);
    19     return 0;
    20 }
    作者:PennyXia
             
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利。
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  • 原文地址:https://www.cnblogs.com/PennyXia/p/12288260.html
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